Ok, I looked at the examples in the book, and it seems pretty straight forward... I first look at f(x,0) to see what happens as we approach along the x-axis. The equation becomes 0/0, which is indeterminate. I did the same thing for f(0,y) to see what happens as we approach along the y-axis, which is also indeterminate. This would mean that the limit doesn't exist, according to the book. I even also approached on a different line, y=x, but that is also indeterminate.

That is the wrong answer however, according to the software. I thought maybe I would have to use l'hopital's rule for limits, but I'm not sure if I am supposed to do that, or if I would need to do it twice (with respect to x AND y). I started to take the partial with respect to x and it got very messy, so I wasn't sure if that was what I needed to do.

This is the very first question we are assigned on limits in calc 3.

EDIT:

I also just tried setting [itex]\sqrt{x^{2}+y^{2}+81}\neq9[/itex] and I got that y cannot be equal to plus or minus x, but I'm not really sure what to do with that.

OHHH. Duh. I see it. I guess I just had to take a screenshot to notice it. Thank you everybody.

</oblivious>

But while I have everyone's attention, I would like some clarification on something.

So I found two paths that agree on a limit, but there are literally an infinite number of paths to approach the point from. How can I be sure every single path agrees on the limit? The way I understand it, even if a single path doesn't agree, then the limit doesn't exist.

How can I be sure that 18 is the correct answer? (it is, according to the software)

But while I have everyone's attention, I would like some clarification on something.

So I found two paths that agree on a limit, but there are literally an infinite number of paths to approach the point from. How can I be sure every single path agrees on the limit? The way I understand it, even if a single path doesn't agree, then the limit doesn't exist.

How can I be sure that 18 is the correct answer? (it is, according to the software)

Good question! Why don't you try a similar manipulation with the new function - clear the denominator and simplify. What do you end up with?